Question: Simplify the following expression and state the condition under which the simplification is valid. $n = \dfrac{2t^2 - 26t + 72}{-8t^3 + 40t^2 - 32t}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ n = \dfrac {2(t^2 - 13t + 36)} {-8t(t^2 - 5t + 4)} $ $ n = -\dfrac{2}{8t} \cdot \dfrac{t^2 - 13t + 36}{t^2 - 5t + 4} $ Simplify: $ n = - \dfrac{1}{4t} \cdot \dfrac{t^2 - 13t + 36}{t^2 - 5t + 4}$ Next factor the numerator and denominator. $ n = - \dfrac{1}{4t} \cdot \dfrac{(t - 4)(t - 9)}{(t - 4)(t - 1)}$ Assuming $t \neq 4$ , we can cancel the $t - 4$ $ n = - \dfrac{1}{4t} \cdot \dfrac{t - 9}{t - 1}$ Therefore: $ n = \dfrac{ -t + 9 }{ 4t(t - 1)}$, $t \neq 4$